Question

How would I solve for x here?
sqrt(3x - 2) + sqrt(11 + x) = 1

Answer 1

square both sides.

Answer 2

Because you have to get rid of that radical, square both sides.

Answer 3

right so.. hold on a sec

Answer 4

\[\rm 3x - 2 + 2(\sqrt{11+x})(\sqrt{3x - 2}) + 11 + x = 1\]

Answer 5

Can you complete it?

Answer 6

i have 4x + 9 + 2sqrt(3x^2 + 31x - 22) = 1

Answer 7

then i guess i have to factor

Answer 8

so I'll just have the sqrt by itself and then put the 4x + 9 on the other side and then I can divide by the 2 in front of the radical to get
sqrt(....) = -2x - 4

Answer 9

Oh yes, you'd have to factor.

Answer 10

so I'm on the right track then?

Answer 11

I'm trying to find a trick here.

Answer 12

No solutions!

Answer 13

that's correct

Answer 14

I solved it and got no solutions as the answer.

Answer 15

how?

Answer 16

I actually raised both sides to power of 3. ;)

Answer 17

But I did it on paper and that was fast...

Answer 18

why did you raise both sides to the third?

Answer 19

I was just trying and testing.

Answer 20

... and then that didn't work.

Answer 21

so I changed my route again.

Answer 22

Let me show my solution here:\[\rm \sqrt{3 x - 2} + \sqrt{x + 11} = 1\]\[\rm 3x - 2 + 2\cdot\sqrt{3x - 2}\cdot \sqrt{x + 11} + x+11 = 1 \]\[\rm 4x + 9 + 2\cdot \sqrt{3x - 2}\cdot \sqrt{x + 11} = 1\]\[\rm 4x + 9 + 2\sqrt{3x^2 + 31x - 22} = 1\]\[\rm 4x + 2\sqrt{3x^2 + 31x - 22} = -8\]\[\rm 2x + \sqrt{3x^2 + 31x - 22}=-8\]

Answer 23

Square both sides again.

Answer 24

Dont need to, isolate the square root an then do it, the square root vanishes

Answer 25

Oh, good point

Answer 26

\[\rm \sqrt{3x^2 + 31x - 22}= -2x - 8\]\[\rm 3x^2 + 31x - 22 = 4x^2 +32x+64\]

Answer 27

I mistakenly squared both sides on paper but it still worked.